Question

Two forces are applied to a tree stump to pull it out of the ground. Force FA has a magnitude of 2200 newtons and points 34.5° south of east, while force FB has a magnitude of 3160 newtons and points due south. Using the component method, find the magnitude and direction of the resultant force FA + FB that is applied to the stump. Specify the direction with respect to due east.

Answer 1

Lon FA C0134.5 34.50 & FA-22oon FA sin 345 FB = 3160N S Component of forces along n-direction Fa= FA cas34.5 Fn= 2200 x 0,824 = 1812,8 N -y direction Component of forces along y direction fy - Fo. FA sin 34,5 =-3160-2200X 0,566 =-3160 1245.2 = 4 4052N 2 Resultant force, F = Jy²fy √( 1812.8) ²+(-4405,2)2

√3 286243.84 + 19405787.04 √ 22 692030.88 4763.615 o = tant Derection of force tant (Fully tent 44052 1812.8 = tant (2,43) o = 67.630 south of east o Fy f YFT